If I have a roller coaster, and hill 1 height is 50cm and hill height 2 is 25 cm and the PE is .9 as it approaches the top of hill 2, how do I calculate the speed.
Please help
THe PE is .9 where? Where did the coaster start?
The coaster started at .18PE-the .9 is the PE at the top of the second hill.The second hill is lower.
Well, the difference in the two Potential energies is now KEnergy, 1/2 mv^2, solve for v.
. The way to calculate this is you take the KE would be .9. You would say .9 = ½ (.035 x v^2) = 2.215. Correct or no?
Should I have done this as 1/2mv^2 = mgh
Cross out like terms and get to v^2 = square root of 2 x g x .25m(h)= 2.125 m/s
Would this be correct?
To calculate the speed at the top of hill 2, you can use the principle of conservation of energy. The total mechanical energy of the roller coaster is the sum of its kinetic energy and potential energy. At the top of hill 2, the roller coaster has no potential energy (as it is at the highest point) and only kinetic energy.
The potential energy (PE) of the roller coaster is given by:
PE = m * g * h
Where m is the mass of the roller coaster, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the hill.
The kinetic energy (KE) of the roller coaster is given by:
KE = 0.5 * m * v^2
Where v is the velocity (speed) of the roller coaster.
Since the total mechanical energy is conserved, we can equate the potential energy at hill 1 to the kinetic energy at the top of hill 2:
PE at hill 1 = KE at hill 2
m * g * h1 = 0.5 * m * v^2
Canceling out the mass (m) from both sides:
g * h1 = 0.5 * v^2
Plugging in the values, with g = 9.8 m/s^2, h1 = 50 cm (or 0.5 m), we can solve for v:
9.8 * 0.5 = 0.5 * v^2
4.9 = 0.5 * v^2
Dividing both sides by 0.5:
9.8 = v^2
Taking the square root of both sides:
v ≈ √9.8
v ≈ 3.13 m/s
Therefore, the speed of the roller coaster at the top of hill 2 is approximately 3.13 m/s.